4Thanks [159]
A non-zero vector a i s parallel to the line of intersection of plane P1 determined by
i+j , i - 2j and plane P2 determined by vector 2i + j ,3i + 2k , then angle between
a and vector i - 2j +2k is
(A)Î /4
(B) Î /2
(C) Î /3
(D) Î
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2 Answers
29See first we need to find the eqn of the normal to the plane P1 by finding the cross product of ( i + j ) and (i - 2j ) which comes out to be -3k
then similarly eqn of the normal to the plane P2 = 2i - 4j - 3k
then to find the eqn of line passing through the intersection of the abv plane we need to find the cross product P1 P2..
which comes out to be12i + 6j..
now find dot product 12i + 6j with i -2j + 2k
cosθ = 0
so θ = π/2...
Ans B